Let $H\subseteq T$ be a maximal Toral subalgebra of a semsimple Lie algebra $L.$ Nowto obtain the so called root space decomposition one proceeds as follows. It is clear that $ad_LH$ consists of mutually commuting elements. My question is why $L$ can be written as $L=\oplus_\alpha L_\alpha$ where $L_\alpha:\{x\in L:[hx]=\alpha(h)x, h\in H\}$ where $\alpha $ ranges over $H^*$? I understand this is a generalization of the usual decomposition of diagonalizable matrix generalized to the case of commuting family of matrices. Bu how to prove this rigorously?


1 Answer 1


I'm assuming here we're talking about Lie Algebras over some an algebraically closed field $\mathsf{k}$, so that a maximal toral subalgebra is abelian. This result then relies on a more general lemma which goes as follows.

Let $\mathfrak{h}$ be a nilpotent Lie algebra with $(V,\rho)$ a finite dimensional representation of $\mathfrak{h}$. Then $V$ decomposes as a direct sum

$$ V = \bigoplus_{\lambda \in \left( \mathfrak{h}/D\mathfrak{h}\right)^{*}} V_{\lambda}, $$


$$ V_{\lambda} = \left\{v \in V \mid \forall \ x \in \mathfrak{h}, \exists n \in \mathbb{N} : \left( \rho(x) - \lambda(x) \right)^n(v) =0 \right\}. $$

To prove this result, we proceed by induction on the dimension of $V$, with the case that the dimension of $V$ is $1$ being trivial.

Suppose now that the dimension of $V$ is greater than $1$, and suppose for the moment that $V$ admits a decomposition of the form $V = U \oplus W$ for some non-zero proper subrepresentations $U,W$, then by the inductive hypothesis then claim holds for $U,W$. But then by voting that $V_{\lambda} = U_{\lambda} \oplus W_{\lambda}$, we would get that the result holds for $V$ too.

Now consider $x \in \mathfrak{h}$, then $\rho(x) : V \to V$ is a linear map, and so $V$ admits a generalised eigenspace decomposition of the form $V = \bigoplus V_{\lambda(x)}$. It can be shown (this is one place where we need that $\mathfrak{h}$ is nilpotent) that each $V_{\lambda(x)}$ is in fact a subrepresentation of $V$, and so if it were the case that $\rho(x)$ had two distinct eigenvalues, then by the above discussion we would be done, and so we may reduce to the case that every $x \in \mathfrak{h}$ has precisely one eigenvalue $\lambda(x)$ say. Now Lie's theorem gives some $0 \neq v \in V$ and a Lie algebra homomorphism $\mu : \mathfrak{h} \to \mathsf{k}$ such that $\rho(y)v = \mu(y)v$ for all $y \in \mathfrak{h}$. But then $\mu(x)$ is an eigenvalue of $\rho(x)$, and so $\mu(x) = \lambda(x)$ for any $x \in \mathfrak{h}$, and so in particular the map $x \to \lambda(x)$ is in fact a member of $(\mathfrak{h} / D\mathfrak{h} )^{*}$. But what we have proven here is precisely that $V = V_{\lambda}$, and so we're done.

Now this is almost the result you want, but we have a bit more work to prove that the $n$s are all equal to $1$ in your case where $\mathfrak{h} = H$, $V = L$ and $\rho = \operatorname{ad}_L$.

One way to see this is to prove that in this case the non-zero $L_{\alpha}$ are all one-dimensional, and this can be proven by a dimension counting argument making use of the killing form, and the $\mathfrak{sl}_{\alpha}$-subalgebra of $L$.

See if you can conclude the proof from here.


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